= Solution
Positivity of the squared norm of the supplied state gives, for every real $s$,
$$
\begin{aligned}
0&\leq\lVert(p-isx)\psi\rVert^2\\
&=\langle(p+isx)(p-isx)\rangle\\
&=\langle p^2\rangle+s^2\langle x^2\rangle
+is\langle[x,p]\rangle\\
&=(\Delta p)^2+s^2(\Delta x)^2-s\hbar,
\end{aligned}
$$
where the stated zero means were used in the last line. This quadratic in $s$ is nonnegative for every real $s$, so its discriminant is nonpositive:
$$
\hbar^2-4(\Delta x)^2(\Delta p)^2\leq0.
$$
This is the <quadratic-norm proof of the Heisenberg uncertainty relation>, and proves
$$
\boxed{\Delta x\,\Delta p\geq\frac\hbar2}.
$$
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